A145515 Square array A(n,k), n>=0, k>=0, read by antidiagonals: A(n,k) is the number of partitions of k^n into powers of k.

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 4, 1, 1, 1, 2, 5, 10, 1, 1, 1, 2, 6, 23, 36, 1, 1, 1, 2, 7, 46, 239, 202, 1, 1, 1, 2, 8, 82, 1086, 5828, 1828, 1, 1, 1, 2, 9, 134, 3707, 79326, 342383, 27338, 1, 1, 1, 2, 10, 205, 10340, 642457, 18583582, 50110484, 692004, 1, 1, 1, 2, 11, 298, 24901, 3649346, 446020582, 14481808030, 18757984046, 30251722, 1, 1

Offset: 0

Author

Alois P. Heinz, Oct 11 2008

Keywords

• nonn
• tabl

Examples

			A(2,3) = 5, because there are 5 partitions of 3^2=9 into powers of 3: [1,1,1,1,1,1,1,1,1], [1,1,1,1,1,1,3], [1,1,1,3,3], [3,3,3], [9].
Square array A(n,k) begins:
  1,  1,   1,    1,     1,      1,  ...
  1,  1,   2,    2,     2,      2,  ...
  1,  1,   4,    5,     6,      7,  ...
  1,  1,  10,   23,    46,     82,  ...
  1,  1,  36,  239,  1086,   3707,  ...
  1,  1, 202, 5828, 79326, 642457,  ...
		

Links

• Alois P. Heinz, Antidiagonals n = 0..40, flattened

Crossrefs

Columns k=0+1, 2-10 give: A000012, A002577, A078125, A078537, A111822, A111827, A111832, A111837, A145512, A145513.

Row n=3 gives: A189890(k+1).

Main diagonal gives: A145514.

Cf. A007318.

Programs

Maple

b:= proc(n, j, k) local nn;
      nn:= n+1;
      if n<0  then 0
    elif j=0  or n=0 or k<=1 then 1
    elif j=1  then nn
    elif n>=j then (nn-j) *binomial(nn, j) *add(binomial(j, h)
                   /(nn-j+h) *b(j-h-1, j, k) *(-1)^h, h=0..j-1)
              else b(n, j, k):= b(n-1, j, k) +b(k*n, j-1, k)
      fi
    end:
A:= (n, k)-> b(1, n, k):
seq(seq(A(n, d-n), n=0..d), d=0..13);

Mathematica

b[n_, j_, k_] := Module[{nn = n+1}, Which[n < 0, 0, j == 0 || n == 0 || k <= 1, 1, j == 1, nn, n >= j, (nn-j)*Binomial[nn, j]*Sum[Binomial[j, h]/(nn-j+h)* b[j-h-1, j, k]*(-1)^h, {h, 0, j-1}], True, b[n, j, k] = b[n-1, j, k] + b[k*n, j-1, k] ] ]; a[n_, k_] := b[1, n, k]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 13}] // Flatten (* Jean-François Alcover, Dec 12 2013, translated from Maple *)

Formula

See program.

For k>1: A(n,k) = [x^(k^n)] 1/Product_{j>=0} (1-x^(k^j)).

Extensions

Edited by Alois P. Heinz, Jan 12 2011